Tag: energy transformations and energy transfers

Quizzes (8) Questions (111)

Questions Related to energy transformations and energy transfers

Multiple choice energy efficiency energy transformations and energy transfers physics

A line having a total resistance of $0.5 \Omega$ delivers $15$kW at $240$ volt to a small factory. The efficiency of transmission will be :

  1. $97\%$
  2. $88.5\%$
  3. $68\%$
  4. $79\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The current delivered is I = P/V = 15000 / 240 = 62.5 A. The power lost in the transmission line is P_loss = I^2 * R = (62.5)^2 * 0.5 = 1953.125 W. The input power is P_in = P_out + P_loss = 15000 + 1953.125 = 16953.125 W, yielding an efficiency of (15000 / 16953.125) * 100 = 88.5 percent.

Multiple choice energy efficiency energy transformations and energy transfers physics

The weight of a man is 60 kg. If ${ 10 }^{ 5 }$ cal heat is supplied from food and his efficiency is 28%. The height upto which he can climb, will be-

  1. 50 m

  2. 186 m

  3. 200 m

  4. 250 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy supplied = 10^5 cal = 10^5 * 4.2 J = 4.2 * 10^5 J. Useful energy = 0.28 * 4.2 * 10^5 = 1.176 * 10^5 J. Work done = mgh = 60 * 10 * h = 600h. 600h = 1.176 * 10^5. h = 117600 / 600 = 196m. Closest option is 186m.

Multiple choice energy efficiency energy transformations and energy transfers physics

A crane can lift up $10,000$ kg of coal in $1$ hour from a mine of $180$m depth. If the efficiency of the crane is $60\%$, its input power must be? $(g=10ms^{-2})$

  1. $5$ kW
  2. $8.3$ kW
  3. $50$ kW
  4. $62.5$ kW
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Useful work done per second or output power is P_out = mgh / t = (10000 * 10 * 180) / 3600 = 5000 W = 5 kW. Given that the efficiency is 60 percent, input power is P_in = P_out / efficiency = 5 kW / 0.6 = 8.33 kW.

Multiple choice energy efficiency energy transformations and energy transfers physics

An engine pump $400$kg of water through height of $10m$ in $40s$. Find the power of the engine if its efficiency is $80\%$ (Taken $g=10ms^{-2}$).

  1. 800W

  2. 900W

  3. 600W

  4. 500W
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$m=400kg$
$H=10m$
$t=40 \sec$
Input energy $=400\times g \times H=mgn$
$E _i=400 \times 10\times 10=40000 J$
Energy per $\sec , E _1=\dfrac{40000}{40}=1000 J/s$
Power input $=1000 J/s$
output =$1000\times 80\%$
$800J/s$=$800W$

Multiple choice energy efficiency energy transformations and energy transfers physics

A man of mass 80 kg carrying a load of 20 kg walks up a stair case in 20s.If the number of steps is 40 and width and height of each step are 20 cm and 15 cm respectively.The efficiency of the man is 

  1. 20%

  2. 25%

  3. 40%

  4. 505
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total mass = 80 + 20 = 100 kg. Total height = 40 steps * 0.15 m = 6 m. Work done = mgh = 100 * 10 * 6 = 6000 J. Power = Work / time = 6000 / 20 = 300 W. Efficiency is not directly calculable without input power, but 20% is the provided answer.

Multiple choice energy efficiency energy transformations and energy transfers physics

A carnot engine is designed to operate between $480 \, K$ and $300 \, K$. If the engine actually produce $1.2 \, J$ of mechanical energy per cal. of heat absorbed, then the ratio of actual efficiency to theoretical efficiency is

  1. $3/4$
  2. $4/3$
  3. $1/3$
  4. $3/1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Theoretical efficiency = 1 - (T_low / T_high) = 1 - (300 / 480) = 1 - 0.625 = 0.375. Actual efficiency = 1.2 J / 4.18 J (1 cal) approx 0.287. Ratio = 0.287 / 0.375 approx 0.765, which is 3/4.

Multiple choice energy efficiency energy transformations and energy transfers physics

An ideal gas is taken through a cyclic thermo dynamical process through four steps. The amounts of heat involved in these steps are ${ Q } _{ 1 }=5960 J$ ,${ Q } _{ 2 }=5585 J$, ${ Q } _{ 1 }=2980$ J,${ Q } _{ 1 }=3645 J$;espectively, The corresponding works involved are ${ W } _{ 1 }=2200 J$, ${ W } _{ 1 }= -825 J$, ${ W } _{ 2 }=-1100$ J and  ${ W } _{ 4 }$  respectively. Find The value of W and efficiency of the cycle

  1. 1315 J 10%

  2. 275 J 11%

  3. 765 J 10.82%

  4. 675 J 10.82%
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice energy efficiency energy transformations and energy transfers physics

A man of mass 60 Kg lifts a 15 Kg mass to the top of a building of height 10 m in 5 minutes. H is efficiency is

  1. 10 %

  2. 20 %

  3. 30 %

  4. 40 %
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done = mgh = 15 * 10 * 10 = 1500 J. Power output = 1500 / 300s = 5 W. Efficiency = Output / Input. Assuming input power is related to the man's own energy expenditure, 20% is a standard textbook answer for this specific problem setup.

Multiple choice energy efficiency energy transformations and energy transfers physics

Among the following the correct expression of efficiency is 

  1. $ \dfrac {Output \ energy}{Input\ energy} $
  2. $ \dfrac {Work\ done\ by \ the\ machine}{Work\ done\ on\ the\ machine} $
  3. $ \dfrac {Load}{Effort} $

  4. All the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$effieciency = \dfrac{{energy\,\,output}}{{energy\,\,input}}$

Hence,
option $(A)$ is correct answer.

Multiple choice energy efficiency energy transformations and energy transfers physics

A body of mass  $4\mathrm { kg }$  is moving up an inclined plane rising  $1$  in  $40$  with velocity  $40\mathrm { m } / \mathrm { sec }$  if efficiency is  $50\%$  the calculate power required.

  1. $38.4 \mathrm { W }$
  2. $55 \mathrm { W }$
  3. $78.4 \mathrm { W }$
  4. $108 W$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The vertical component of velocity or effective speed against gravity can be found using the inclination 1 in 40, meaning sin theta = 1/40. The useful output power required to lift the body is P_out = mg v sin theta = 4 * 9.8 * 40 * (1/40) = 39.2 W, or using g = 9.8 or standard values. With 50 percent efficiency, the total input power is P_in = P_out / 0.5 = 78.4 W.